SOLUTION: Tutors I submitted this once showing my work but I can't find my notes. If you sear your email you will see how I worked the problem. I asked if you could check my work. Here is
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Question 144422: Tutors I submitted this once showing my work but I can't find my notes. If you sear your email you will see how I worked the problem. I asked if you could check my work. Here is the word problem again..Thank you $5000 is invested at interest rate K, compounded continuously and grows to $6954.84 in 6 years. Find the interest rate; the exponential growth function; the balance after 10 years and find the doubling time.
If I remember my notes I think I had the doubling time as 12 years, the interest rate at 6%.
Thank you Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! $5000 is invested at interest rate K, compounded continuously and grows to $6954.84 in 6 years. Find the interest rate; the exponential growth function; the balance after 10 years and find the doubling time.
:
P*e^rt = A
:
5000*e^6r = 6954.84
e^6r =
Divide both sides by 5000
e^6r = 1.390968
:
Find the nat log of both sides:
ln(e^6r) = 1.390968
:
log equiv of exponents
6r*ln(e) = ln(1.390698)
:
Find the ln of both sides (ln(e) = 1)
6r = .33000
r =
r = .055, 5.5%
:
Growth function
f(t) = 5000*e^.055t
:
use this to check above solution on a calc: enter: 5000*e^(.055*6) = 6954.84
:
:
the balance after 10 years
A = 5000*e^(.055*10)
A = 8666.26
:
and find the doubling time.
1*e^.055t = 2
.055t*ln(e) = ln(2)
.055t = .693147
t =
t = 12.6 yrs
:
Check this using the value of 5000 on calc: enter 5000*e^(.055*12.6) = 9998.53 ~ 10000
